What is the Integral of #sec^4 (x) tan^4 (x) dx#?

Question

Ezra Adams · Accepted Answer

This is one of those problems where you just have to improvise to see what you can come up with.

Two ways I can think of to do this:

I already tried the first way and it didn't go well because both trig functions had the same exponent, so here is the second way.

#int sec^4xtan^4xdx#

Separate to achieve #sec^2x# as a product term.

#= int sec^2xsec^2x(tanx)^4dx#

Now transform #secx# terms that aren't the #sec^2x# you want to preserve to achieve #tanx# terms via the trig relationship #sec^2x = tan^2x + 1#.

#= int sec^2x (tan^2x + 1)(tanx)^4dx#

Now that we have #tanx# terms we can isolate as #u#... Let #u = tanx#. Then, #du = sec^2xdx#.

#=> int (u^2 + 1)(u^4)du#

#= int u^6 + u^4du#

#= u^7/7 + u^5/5#

Pretty much done now. Substitute #u = tanx# back in to get:

#= color(blue)(tan^7x/7 + tan^5x/5 + C)#

And just to check that this worked...

#d/(dx)[tan^7x/7 + tan^5x/5 + C]#

#= tan^6x*sec^2x + tan^4x*sec^2x# (chain rule)

#= sec^2x(tan^6x + tan^4x)# (factor)

#= sec^2x(tan^4x(tan^2x + 1))# (factor)

#= sec^2x(tan^4xsec^2x)# (trig identity)

#= color(green)(sec^4xtan^4x)# (associate/distribute)

Isabella Ackerman · Answer

undefined To find the integral of sec^4(x) tan^4(x) dx, you can use trigonometric identities to simplify the expression. One approach is to rewrite sec^4(x) as (sec^2(x))^2 and tan^4(x) as (sec^2(x) - 1)^2. Then, you can use the substitution method or integrate by parts to solve the integral.

Isabella Ackerman · Answer

undefined To find the integral of $ \sec^4(x) \tan^4(x) \, dx $, you can use the trigonometric identity $ \sec^2(x) = 1 + \tan^2(x) $ to rewrite the integral in terms of $ \sec^2(x) $:

$$ \sec^4(x) \tan^4(x) \, dx = (\sec^2(x) \tan^2(x)) \cdot (\sec^2(x) \tan^2(x)) \, dx $$

Then, substitute $ \sec^2(x) = 1 + \tan^2(x) $:

$$ = ((1 + \tan^2(x)) \tan^2(x)) \cdot ((1 + \tan^2(x)) \tan^2(x)) \, dx $$

This simplifies to:

$$ = (\tan^2(x) + \tan^4(x)) \cdot (\tan^2(x) + \tan^4(x)) \, dx $$

Expand this expression:

$$ = \tan^4(x) + 2\tan^6(x) + \tan^8(x) \, dx $$

Now, you can integrate each term separately:

$$ \int \tan^4(x) \, dx = \frac{1}{5} \tan^5(x) + C $$

$$ \int 2\tan^6(x) \, dx = \frac{2}{7} \tan^7(x) + C $$

$$ \int \tan^8(x) \, dx = \frac{1}{3} \tan^3(x) - \tan(x) + C $$

Combine these results:

$$ \int \sec^4(x) \tan^4(x) \, dx = \frac{1}{5} \tan^5(x) + \frac{2}{7} \tan^7(x) + \frac{1}{3} \tan^3(x) - \tan(x) + C $$